Answer:
<u>One lone-Pair is present in Ammonia</u>
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Explanation:
The number of valence electron in N = 5
The number of Valence electron in H = 1
The formula of ammonia = NH3
Total valence electron in ammonia molecule = 5 +3(1) = 5+3 = 8
The lewis structure suggest that :
Nitrogen completes its octet by sharing the electron pair with 3 hydrogen atoms.
3 electron of Nitrogen are involved in sharing with Hydrogen
So,<u><em> remaining two electron are left non-bonded</em></u> . Hence they exist as lone- pair
So, there is only 1 lone pair in the ammonia molecule .
The shape of NH3 is bent according to VSEPR theory . This is so because the presence of 1 lone pair causes more repulsion and occupy more space.
Thus the lone pair is changing the shape of the ammonia molecule . It also increase the dipole moment of the molecule , which gives polarity to it.
Explanation:
Magnesium has atomic number 12. It will be distributed in K, L, M shell in the following way:
K shell can accommodate a maximum of 2 electrons.
L shell can accommodate a maximum of 8 electrons and
M shell will accommodate 2 electrons
So the configuration becomes 2, 8,3
Answer:
Kp = 0.049
Explanation:
The equilibrium in question is;
2 SO₂ (g) + O₂ (g) ⇄ 2 SO₃ (g)
Kp = p SO₃² / ( p SO₂² x p O₂ )
The initial pressures are given, so lets set up the ICE table for the equilibrium:
atm SO₂ O₂ SO₃
I 3.3 0.79 0
C -2x -x 2x
E 3.3 - 2x 0.79 - x 2x
We are told 2x = partial pressure of SO₃ is 0.47 atm at equilibrium, so we can determine the partial pressures of SO₂ and O₂ as follows:
p SO₂ = 3.3 -0.47 atm = 2.83 atm
p O₂ = 0.79 - (0.47/2) atm = .56 atm
Now we can calculate Kp:
Kp = 0.47² /[ ( 2.83 )² x 0.56 ] = 0.049 ( rounded to 2 significant figures )
Note that we have extra data in this problem we did not need since once we setup the ICE table for the equilibrium we realize we have all the information needed to solve the question.
Answer:
HCl is the correct answer
